prmo solution 20.08 - wordpress.com€¦ · · 2017-08-202017-08-20 · suppose ab = 6, cd = 8....

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PRMO Solution

20.08.2017

1. How many positive integers less than 1000 have the property that the sum of the digits of each such

number is divisible by 7 and the number itself is divisible by 3?

2. Suppose a, b are positive real numbers such that a a b b 183 . a b b a 182 . Find

9

a b5

3. A contractor has two teams of workers: team A and team B. Team A can complete job in 12 days and

team B can do the same job in 36 days. Team A starts working on the job and team B joins team A after

four days. The team A withdraws after two more days. For how many more days should team B work to

complete the job?

4. Let a, b be integers such that all the roots of the equation 2 2x ax 20 x 17x b 0 are negative

integers. What is the smallest possible value of a + b?

5. Let u, v, w be real numbers in geometric progression such that u v w . Suppose 40 n 60u v w .

Find the value of n?

6. Let the sum

9

n 1

1

n n 1 n 2

written in its lowest term be

p

q. Find the value of q – p.

7. Find the number of positive integers n. Such that n n 1 11 .

8. A pen costs Rs 11 and a notebook costs Rs 13. Find the number of ways in which a person can spend

exactly Rs 1000 to buy pens and notebooks.

9. There are five cities A, B, C, D and E on a certain island. Each city is connected to every other city by

road. In how many ways can a person starting form city A come back to A after visiting some cities

without visiting a city more than once and without taking the same road more than once? (The order in

which he visits the cities also matters: e.g., the routes A B C A and A C B A are

different)

10. There are eight rooms on the first floor of a hotel, with four rooms on each side of the corridor,

symmetrically situated (that is each room is exactly opposite to one other room). Four guests have to be

accommodated in four of the eight rooms (that is, one in each) such that no two guests are in adjacent

rooms or in opposite rooms. In how many ways can the guests be accommodated?

11. Let x 3x

f x sin cos3 10

for all real x. Find the least natural number n such that f n x f x for

all real x.

12. In a class, the total numbers of boys and girls are in the ratio 4 : 3. On one day it was found that 8 boys

and 14 girls were absent from the class, and that the number of boys was the square of the number of

girls. What is the total number of students in the class?

13. In a rectangle ABCD, E is the midpoint of AB; F is a point on AC such that BF is perpendicular to AC,

and FE perpendicular to BD. Suppose BC = 8 3 . Find AB.

14. Suppose x is a positive real number such that x , x andx are in a geometric progression. Find the

least positive integer n such thatnx 100 . (Here [x] denotes the integer part of x and {x} = x – [x].)

15. Integers 1, 2, 3, …., n. Where n > 2, are written on a board. Two numbers m, k such that

1 < m < n, 1 < k < n are removed and the average of the remaining numbers is found to be 17. What is

the maximum sum of the two removed numbers?

16. Five distinct 2-digit numbers are in a geometric progression. Find the middle term.

17. Suppose the altitudes of a triangle are 10, 12 and 15. What is its semi-perimeter?


18. If the real numbers x, y, z are such that 2 2 2x 4y 16z 48 and xy 4yz 2zx 24 . What is the

value of 2 2 2x y z ?

19. Suppose 1, 2, 3 are the roots of the equation4 2x ax bx c . Find the value of c.

20. What is the number of triples (a, b, c) of positive integers such that (i) a < b < c < 10 and (ii) a, b, c, 10

form the side of a quadrilateral?

21. Find the number of ordered triples (a, b, c) of positive integers such that abc = 108.

22. Suppose in the plane 10 pair-wise nonparallel lines intersect one another. What is the maximum possible

number of polygons (with finite areas) that can be formed?

23. Suppose an integer x, a natural number n and a prime number p satisfy the equation2 n7x 44x 12 p

. Find the largest value of p.

24. Let P be an interior point of a triangle ABC whose side lengths are 26, 65, 78. The line through P

parallel to BC meets AB in K and AC in L. The line through P parallel to CA meets BC in M and BA in

N. The line through P parallel to AB meets CA in S and CB in T. If KL, MN, ST are of equal lengths,

find this common length.

25. Let ABCD be a rectangle and let E and F be points on CD and BC respectively such that area (ADE) =

16, area (CEF) = 9 and area (ABF) = 25. What is the area of triangle AEF?

26. Let AB and CD be two parallel chords in a circle with radius 5 such that the centre O lies between these

chords. Suppose AB = 6, CD = 8. Suppose further that the area of the part of the circle lying between the

chords AB and CD is m n / k , where m, n, k are positive integers with gcd m,n,k 1 . What is

the value of m + n + k?

27. Let 1 be a circle with center O and let AB be a diameter of 1 . Let P be a point on the segment OB

different from O. Suppose another circle 2 with centre P lies in the interior of 1 . Tangents are drawn

from A and B to the to the circle 2 intersecting 1 again at A1 and B1 respectively such that A1 and

B1 are on the opposite sides of AB. Given that A1B = 5. AB1 = 15 and OP = 10, find the radius of 1 .

28. Let p,q be the prime numbers such that n3pq – n is a multiple of 3pq for all positive integers n. Find the

least possible value of p + q.

29. For each positive integer n, consider the highest common factor hn of the two numbers n! + 1 and (n +

1)!. For n < 100, find the largest value of hn…

30. Consider the areas of the four triangles obtained by drawing the diagonals AC and BD of a trapezium

ABCD. The product of these areas taken two at time are computed. If among the six products so

obtained, two products are 1296 and 576, determine the square root of the maximum possible area of the

trapezium to the nearest integer.


Solutions 1. Considering both the conditions , the sum of digits must be divisible by 21. The below table

consists of

Sequences of digits whose sum is 21.

Sequence Number of Permutations

(3, 9, 9) 2

(4, 8, 9) 6

(5, 7, 9) 6

(6, 6, 9) 2

(5, 8, 8) 2

(5, 9, 8) 3

(6, 7, 8) 3

(6, 9, 7) 3

(4, 9,7) 1

= 28

2. a a b b 183

a b b a 182

a u ; b v

Solving

3 3u v 183

2 2u v v u 182

3 3u v 183 ........ i

2 23u v 3v u 3 182

2 23u v 3v u 546 ......... ii

3 3 2 2u v 3u v 3v u 729

3

u v 729

3 3

u v 9

u v R

u v 9

uv (u + v) = 182

uv = 182

9

182

u 9 u9

9u 9 u 182

281u 9u 182


29u 81u 182 0

Solve to get 196 169

a , b9 9

9 a b 9 365

735 1 5 9

3. Team A complete the job in 12 days

1 day work of Team A = 1

12

Team B complete the job in 36 days

1 day work of Team B = 1

36

(A+ B) team 1 day work = 1 1 3 1 1

12 36 36 9

A’s 4 days work = 4 1

12 3

Work left = 1 2

13 3

Team (A+B) 2 days work = 2

9

Work left = 2 2 6 2

3 9 9

=

4

9

Team B do 1

36work in 1 day

Team B do 4

9work in =

36 4

9

= 16 days

4. 2 2(x a x 20)(x 17x b) 0

All roots are –ve smallest value (a+b) =?

(x+ ) (x+ )(x+ ) (x+ )=0

Here + =a here + =17

= 20 = b

+ =a

Vice versa 1 20 21

as & 2 10 12

4 5 9min

=4 = 5

Vice Versa =b

as & 1 16 16min

2 15 30>16

3 14 >16

4 13 >16

5 12 >16

6 11 >16


7 10 >16

8 9 >16

For min (a+b) we need min (a) & min (b)

min (a+b) = 9+16 = 25 5. u = u

v = ur w = ur2 u10 = un rn = u60 r120 r120 = u-20

r = 1

6u

u40 = un .n

6u

= 5n

6u

40 = 5n

6

n = 40.6

5= 48

6. 9

n 1

1

n(n 1)(n 2)

1

n(n 1)(n 2) =

1

2n+

1

(n 1)

+

1

2(n 2)

9

n 1

1 1 1 1 1 1 1 1 1 1

2n 2 4 6 8 10 12 14 16 18

9

n 1

1 1 1 1 1 1 1 1 1 1

(n 1) 2 3 4 5 6 7 8 9 10

9

n 1

1 1 1 1 1 1 1 1 1 1

2(n 2) 6 8 10 12 14 16 18 20 22

9

n 1

1 1 1 1 1 1 1 1

n(n 1)(n 2) 16 5 8 10 16 20 22

= 1 1 1

4 20 22

= 1 1

5 22

= 27

110

110-27

= 83

7. n n 1 11 n I


both sides are +ve

squaring both sides

n + n + 1 + 2 n(n 1) < 121

2n+ 2 n(n 1) < 120

n n(n 1) < 60

n(n 1) is always > n but < n+1

in decimal representation

n+ n+ fraction < 60

2n + fraction < 60

n can befrom1to 29

as58 fraction 59 60

29 Values of X

8. 11P + 13N = 1000

1000 13N

P11

Maximum possible value for N can be 76. Possible values for N for the above equation is in a

pattern

N = 5, 16, 27, 38, 49, 60, 71

9. When we go through 1city we need to follow

The Same path back

No of ways = 0

when through 2 cities we have 4p2 ways i.e 12

when through 3 cities 4p3 = 24 when

when through 4 cities 4p4 = 24

Total we of ways = 0+12+24+24 = 60


10. Case i Case ii

No of ways = 4! No of ways = 4!

Total = 2x4!

= 48ways

11. f(x)= sin x

3+ cos

3x

10

Period of sin x

3= 6

Period of cos 3x

10=

20

3

Period of sin x

3+ cos

3x

10= LCM of 6 ,

20

3

= LCMof 6 &20

HCFof 1&3

= 60

1

= 60

f ( 60 + x) = f(x)

n= 60

12. Number of boys = 4x

Number of girls = 3x

According to question

2

4x 8 3x 14

4x – 8 = 9x2 + 196 – 84x

20 9x 196 84x 4x 8

20 9x 88x 204


88 7744 7344

x18

88 400

x18

88 20

x18

108 68x 16 or (neglect)

10 18

x=6

boys = 4x = 24

girls = 3x = 18

total = 42

13.

in rt d AFB

AF = 2y coa ;

BF = 2y sin ;

In rt d BGE

BG2 + GE2= BE2

(ysin )2 + (2y sin )2 = (sin2 )2 = y2

Sin2 +4sin2 sin22 = y 0

Cos2 = 4sin2 sin22 [ sin2 a = 2 sin a cos a ]

16 sin4 =1

(sin )4 =

41

2

sin = 1

2 = 300

tan = BC 8 3 4 3 1

AB 2y y 3 4 x 3 = y

y = 12


AB = 2Y = 212= 24 cm

14. given { x } [ x ] x are in GP xR+

least +ve integer S.T xn > 100

x = { x } r2

[ x ] = { x } r

r2 = r+1

r = 1 5

2

(but it can’t be (-ve) as x is + ve always [ x ] can’t be (-ve) )

r= 1 5

2

1.61

As 0 < { x } < 1 if { x } = 0 [ x ] = 0

0 < [ x ] < r then x =0

0 < [ x ] < 1 5

2

x 0 as 0n > 100 for n

[ x ] = 1

{ x } = 1

r=

2

1 5

x = 21

rr

= 1 5

r2

n

1 5

2

> 100

n log 1 5

2

> 2

n > 2

log1.61

2

log1.61<

2

log1.6

2

log1.61<

200

0.20

2

log1.61< 10

nearest + ve integer >2

log1.61is n = 10


15.

n(n 1) n(n 1)(2n 1) 3

2 217n 2 n 2

2 2n n 4n 2 n n 6

172(n 2) 2(n 2)

2n 3n 2 (n 3)(n 2)

172(n 2) 2(n 2)

n 1 n 3

172 2

n < 35 and n > 31

n = 32, 33, 34

case-1, n =32

n(n 1)p

2 17(n 2)

n(n 1)17 (n 2) p

2

P =18

Case-2, n =33 P= 34

Case-3, n= 34 p = 51

Maximum sum = 51

16. Let the terms be a, ar, ar2, ar3, ar4

Each of the terms is a 2-digit no.

Possible values for ‘r’ are 2

3or

3

2

Case 1 a= 16, r =3

2

a = 16

ar = 3

162

= 24

ar2 = 36

Case 2 a = 81, r =2

3

a = 81

ar2 = 4

819

= 36

Case 3 a = 81, r 3

2

Values are not 2 digit no. similarly


Case 4 a = 16 r= 2

3

Discard

17.

1 1 1

10 a 12 b 15 c2 2 2

10a 12b 15c

L.C.M of 10,12,15 60

Dividing by 60

a b c

6 5 4

Let a b c

k6 5 4

a 6k,b 5k,c 4k

s(s a)(s b)(s c)

15k 15k 15k 15k

6k 5k 4k2 2 2 2

15k 3k 5k 7k

2 2 2 2

215 7k

4

21 15k 76k 10

2 4

8

k7

48

a7

, 40

b7

32

c7


120

a b c7

a b c 60

2 7

18. x2 + 4y2 + 16z2 = 48

By priniciple of home geneity let all 3 terms be equal

x2 = 4y2 = 16z2 = 16

x= 4 y = 2 Z= 1

xy + 4yz + 2zx = 24

Put x =4 y= 2 z = 1 here

xy = 4yz = 2zx =8

It also follows principle of homogeneity

x2+y2+z2 = 42+22+12 = 21

19. x4 + 0.x3 + ax2 + bx = c

1 + 2 + 3 + 8 = 0

1

= 0

= -6

1.2.3. = -36 = -c

= 36

20. a, b, c,

Where

a+b+c>10 a < b< c < 10

We start with assuming a =1, b = 2, then using a +b +c > 10

c >7

i.e

c = 8, 9

no of cases = 2

i.e

(1, 2, 8) or (1, 2, 9)

Now keep a = 1, increase ‘b’ by 1 ‘b’ become = 3

C > 6

No. of cases = 3

i.e (1, 3, 7) ; ( 1, 3, 8); (1, 3, 9)

when we continue in the same fashion we get the following we get the following table


a b c No of

cases

1 2 8,9 2

1 3 7,8,9 3

1 4 6,7,8,9 4

1 5 6,7,8,9 4

1 6 7,8,9 3

1 7 8,9 2

1 8 9 1

2 3 6,7,8,9 4

2 4 5,6,7,8,9 5

2 5 6,7,8,9 4

2 6 7,8,9 3

2 7 8,9 2

2 8 9 1

3 4 5,6,7,8,9 5

3 5 6,7,8,9 4

3 6 7,8,9 3

3 7 8,9 2

3 8 9 1

4 5 6,7,8,9 4

4 6 7,8,9 3

4 7 8,9 2

4 8 9 1

5 6 7,8,9 3

5 7 8,9 2

5 8 9 1

6 7 8,9 2

6 8 9 1

7 8 9 1

21. 108 = ab c

108 = 2233

If N= xa yb

Total solution = a+2C2 b+2C2

total solution

4C2 5C2

= 6x10 = 60


22. No of nm- overlapping polygons

= 2n 3n 2

2

= 2(10) 3 10 2

2

= 100 30 2

2

= 36

23. (7x-2) (x-6)=pn

7x-2= p and x -6 = p

40 = p - 7 p

If , N, p is a divisor of 40 p = 2 or 5

If p =2 , 40 = 2 - 7.2

23.5 = 2 - 7.2

= 3 and 2 = 40+56 Z hence not possible

If p =5 then 40 = 5

- 7.5

32 .5 5 7.5

= 1 and 5

= 40 +35 z hence not possible

So = 0 p = 47 p =47 and = 1

24.

k a + l a + ma = a

Common length = a ( k +m ) =b ( k + l ) = c ( l + m )

a( 1-l ) = b (1-m) = c ( l +m )

65 (1 – l) = 78 (l – m ) = 26 ( l + m)

5 (1 –l) = 6( l-m ) = 2 (l+m)

5-5l = 2l+2m 5-5l = 6- 6m

7l + 2m = 5 6m-5l -1

6-6m = 2l +2m

2l+8m = 6

(i) 26l = 14

7

l13

m= 5 7l 8

2 19


common length = 26 15

3013

25.

ab = 32 cd = 18 ef = 50

F = b+c a = e+d

Possible values

a = 1, 32, 4, 8 = b

c = 1, 2, 9, 3, 6, 18 =d

e = 1, 50, 2, 25, 5, 10 =f

only

a = 8 and f = 10 are satisfying the condition

area of rectangle = 810 = 80

ar (AFE) = 80- 50 = 30 units

26.

here clearly

AOB = 740

COD = 1060

AOC + BOD = 1800

Area bw

chord AB & CD

= Ar (AOB) + Ar ( COD) + Ar (AOCA)+Ar(BODB)

= 21 1 5

.4.6 .3.82 2 2

= 24 + 25

2


= 48 25

2

=

m n

k

m=48 n = 25 k = 2

m+k+n=75

27.

Here AA’B = 900 AB’B = 900

In AB’B in AA’B

AB’B ~ PXB AA’B~ AYP

AB' AB

PX PB

A 'B AB

YP AP

15 2R

r R 10

5 2R

r R 10

= 15(R 10) 5(R 10)

2rR R

3(R-10) = R+10

2R = 40

R = 20 unit

28. 28

29. n! + 1is not divisible by 1,2 ………,n

(n+1) ! is divisible by 1, 2,………,n

So HCF n+1

Also (n+1)! Is not divisible by n +2, n+3….

Let n =99

99! 1 and 100!

HCF = 100 is not possible as 100 divides 99!

Not possible with composites

consider prime i.e. n= 97

Now 96! +1 and 97! Are both divisible by 97

So HCF = 97


30.

Case 1 a2 k = 576 - 1

a2 k3 = 1296 - 2

2 3

2

a k 1296

a k 576

2 9k

4

3k

2

( neglect –ve)

a2 = 2

5763

a2 = 192 x 2

a2 = 384

Case 2 2 2

2

a k

ak=

1296

576

K= 9

4

A = 16

Case3 a2k3 = 1296 a2k2= 576

k2 = 1296

576

k= 9

4

a = 32

3

area = a (k+1)2

1area 169 13

2

2

3area 1 122.5

2

2

3

32 9area 1

3 4

prmo solution 20.08 - wordpress.com€¦ · · 2017-08-202017-08-20 · suppose ab = 6, cd = 8....

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